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Q.
Total number of geometrical isomers shown by compounds $(A)$ is__________.
Organic Chemistry – Some Basic Principles and Techniques
Solution:
In this case, terminal groups are same.
No. of double bonds $n=2$.
No. of $G.I. =2^{n}-1+2^{n / 2-1}$
$=2^{1}+2^{0}=3$
Following are the $3$ structures:
But (iii) and (iv) are same structures. Hence, $3$ geometrical isomers are possible.
